Suppose $X$ is a normal space and $f:X \rightarrow X$ is continous on X, and also uniformly continuous on a subset $A \subseteq X$. In this setting, can one conclude that f is uniformly continuous on the closure of $A$?
If not, what about a metric space? Other families of spaces?
I ask this question for the following case:
Montel's Theorem in complex analysis states that a bounded sequence of holomorphic functions on an open set has a converging subsequence (i.e, the set of bounded holomorphic functions is sequentially compact and thus compact).
In the proof, one first shows that there is a subsequence that is uniformly continous on a dense subset, and than you extend it to the whole set. The proof of the last assertion heavily uses properties of holomorphic functions (cauchy's theorems and so on). However, it seems to me that it can be proved in a (much) more general case.
Any ideas?
Best Answer
I'll assume that $X$ is a metric space.
Fix $\epsilon > 0$. Let $\delta > 0$ be a constant that satisfies $d(f(a), f(b)) < \epsilon$ for all $a, b \in A$ when $d(a, b) < \delta$. This is possible since $f$ is uniformly continuous on $A$.
Let $x, y \in \overline A$ such that $d(x, y) < \delta / 3$. Since $f$ is continuous at $x$ and $y$, we can choose $\hat x, \hat y \in A$ such that $d(x, \hat x) < \delta / 3$, $d(y, \hat y) < \delta / 3$, $d(f(x), f(\hat x)) < \epsilon/3$ and $d(f(y), f(\hat y)) < \epsilon / 3$.
We have $$ d(f(x), f(y)) \le d(f(x), f(\hat x)) + d(f(\hat x), f(\hat y)) + d(f(y), f(\hat y)) < \epsilon $$ as desired.
It is worth noting that if the codomain of $f$ is complete (which seems to be true in the special case you're considering), then it is sufficient to assume that $f$ is defined and uniformly continuous on $A$. In this case, $f$ has a unique extension to $\overline A$ and the extension is uniformly continuous. See for example this question.